Chapter 5.6, Problem 21E

a.

To determine

To find: the speed of the boat approaching the dock when 10 feet of rope are out.

Answer to Problem 21E

The speed of the boat is $-2.5\text{ feet/sec}$ .

Explanation of Solution

Given information:

From the given diagram in the question we can see that,

  $h=6\text{ feet}$ .

  $r=10\text{ feet}$ .

  $\frac{dr}{dt}=-2.5\text{ feet/sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the speed of the boat approaching the dock when 10 feet of rope are out.

Therefore, from right angled triangle,

Let x be the distance of the boat from dock.

  $\begin{array}{l}{r}^2=h^2+x^2\\ x^2=r^2-h^2\\ x=\sqrt{r^2-h^2}\end{array}$ .

Put the value of $h=6$ in above equation.

  $\begin{array}{l}x=\sqrt{r^2-6^2}\\ x=\sqrt{r^2-36}\\ x={\left(r^2-36\right)}^{\frac{1}{2}}\end{array}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{dx}{dt}=-\frac{1}{2}{\left[r^2-36\right]}^{\frac{1}{2}-1}·\left(2r\frac{dr}{dt}\right)\\ \frac{dx}{dt}=-\frac{1}{2}{\left[r^2-36\right]}^{-\frac{1}{2}}·\left(2r\frac{dr}{dt}\right)\\ \frac{dx}{dt}=-\frac{1}{2\sqrt{r^2-36}}·\left(2r\frac{dr}{dt}\right)\end{array}$

Putting the value of $r=10$ and $\frac{dr}{dt}=\left(-2\right)$ ,

  $\begin{array}{l}\frac{dx}{dt}=\frac{10}{\sqrt{{10}^2-36}}·\left(-2\right)\\ \frac{dx}{dt}=\frac{10}{\sqrt{100-36}}·\left(-2\right)\\ \frac{dx}{dt}=\frac{10}{\sqrt{64}}·\left(-2\right)\\ \frac{dx}{dt}=\frac{10\left(-2\right)}{8}\\ \frac{dx}{dt}=\frac{-20}{8}\\ \frac{dx}{dt}=-2.5\text{ feet/sec}\text{.}\end{array}$

Hence, the speed of the boat is $-2.5\text{ feet/sec}$ .

b.

To determine

To find: the rate angle changing at that moment.

Answer to Problem 21E

The value of $\frac{d\theta }{dt}$ is $-0.15\text{ radians/sec}$ .

Explanation of Solution

Given information:

From the given diagram in the question we can see that,

  $h=6\text{ feet}$ .

  $r=10\text{ feet}$ .

  $\frac{dr}{dt}=-2.5\text{ feet/sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the rate angle changing at that moment.

Therefore,

From right angled triangle

  $\begin{array}{l}\cos \theta =\frac{\text{base}}{\text{hypotenuse}}=\frac{h}{r}=\frac{6}{10}\\ \theta =\cos -\left(\frac{6}{10}\right)\\ \theta =53.1\end{array}$ ,

Again,

  $\cos \theta =\frac{6}{r}$ .

Differentiate with respect to t .

  $\begin{array}{l}-\sin \theta \frac{d\theta }{dt}=-6r^{-2}\frac{dr}{dt}\\ \frac{d\theta }{dt}=\frac{6}{r^2\sin \theta }·\frac{dr}{dt}\end{array}$

Putting the value of $r=10$ , $\theta =53.1$ and $\frac{dr}{dt}=\left(-2\right)$ in above equation.

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{6}{{\left(10\right)}^2\sin 53.1}·\left(-2\right)\\ \frac{d\theta }{dt}=\frac{-12}{100\times 0.8}\\ \frac{d\theta }{dt}=\frac{-12}{80}\\ \frac{d\theta }{dt}=-0.15\text{ radians/sec}\end{array}$

Hence, the value of $\frac{d\theta }{dt}$ is $-0.15\text{ radians/sec}$ .